# Probability Aptitude Questions and Answers:

#### Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Probability Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

1. Two coins are tossed. What is the probability of getting atleast one head?

1. $$\frac{4}{3}$$
2. $$\frac{3}{4}$$
3. $$\frac{1}{2}$$
4. $$\frac{2}{3}$$

Answer: (b) $$\frac{3}{4}$$

Solution: total number of events = $$4$$

Probability that no head occurs = $$\frac{1}{4}$$

then probability that atleast one head occurs $$= 1 - \frac{1}{4} = \frac{3}{4}$$

1. From the pack of $$52$$ playing cards, two cards are picked at random. What is the probability that both the cards are kings?

1. $$\frac{1}{22}$$
2. $$\frac{1}{220}$$
3. $$\frac{1}{221}$$
4. $$\frac{1}{222}$$

Answer: (c) $$\frac{1}{221}$$

Solution: total number of cards = $$52$$

number of kings = $$4$$, then $$P = \frac{^4C_2}{^{52}C_2}....(1)$$ where as $$^4C_2 = \frac{4 \times 3}{2} = 6$$ $$^{52}C_2 = \frac{52 \times 51}{2} = 1326$$ by putting these values in equation $$(1)$$, $$P = \frac{6}{1326} = \frac{1}{221}$$

1. Three coins are tossed. What is the probability of getting two heads?

1. $$\frac{1}{2}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{4}$$
4. $$\frac{1}{5}$$

Answer: (a) $$\frac{1}{2}$$

Solution: sample space = $$\{HHH, \ HHT, \ HTT, \ HTH\}$$ $$\{TTT, \ THH, \ TTH, THT\}$$

total number of events = $$2^3$$ = $$8$$

number of favourable outcomes = $$4$$

then probability of getting two heads, $$P = \frac{4}{8} = \frac{1}{2}$$

1. If five coins are thrown, then find the probability of getting atleast one tail?

1. $$\frac{32}{31}$$
2. $$\frac{31}{32}$$
3. $$\frac{1}{32}$$
4. $$\frac{1}{31}$$

Answer: (b) $$\frac{31}{32}$$

Solution: total number of events = $$2^5$$ = $$32$$

probability that no tail occurs = $$\frac{1}{32}$$

then probability that atleast one tail occurs, $$= 1 - \frac{1}{32} = \frac{31}{32}$$

1. A bag contains $$6$$ white, $$7$$ black and $$8$$ Red balls. If a ball is drawn at random, then what is the probability that it is not black?

1. $$\frac{2}{3}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{2}$$
4. $$\frac{3}{4}$$

Answer: (a) $$\frac{2}{3}$$

Solution: the probability that the ball is of black colour = $$\frac{7}{21}$$ = $$\frac{1}{3}$$

then the probability that the ball is not of black colour = $$1 - \frac{1}{3}$$ = $$\frac{2}{3}$$

1. If three coins are thrown, then find the probability of not getting head?

1. $$\frac{1}{2}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{6}$$
4. $$\frac{1}{8}$$

Answer: (d) $$\frac{1}{8}$$

Solution: total number of possible outcomes = $$8$$

there is only one way of not getting head, so

favourable outcome = $$1$$, then $$P = \frac{1}{8}$$

1. A bag contains $$15$$ Red and $$18$$ Green balls. If one ball is drawn at random, then what is the probability that it is not Red?

1. $$\frac{6}{12}$$
2. $$\frac{6}{13}$$
3. $$\frac{6}{11}$$
4. $$\frac{6}{10}$$

Answer: (c) $$\frac{6}{11}$$

Solution: the probability that the ball is of Red colour = $$\frac{15}{33}$$ = $$\frac{5}{11}$$

the probability that the ball is not of Red colour, $$= 1 - \frac{5}{11} = \frac{6}{11}$$

1. A fair dice is thrown, What is the probability of getting a odd number?

1. $$\frac{1}{2}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{4}$$
4. $$\frac{1}{5}$$

Answer: (a) $$\frac{1}{2}$$

Solution: number of dice = $$\{1, \ 2, \ 3, \ 4, \ 5, \ 6\}$$

odd numbers on dice = $$\{1, \ 3, \ 5\}$$

number of possible outcomes = $$6$$

number of favourable outcomes = $$3$$, then $$P = \frac{3}{6} = \frac{1}{2}$$

1. A bag contains $$3$$ Red, $$4$$ Black balls and an other bag contains $$5$$ Red, $$6$$ Black balls. If one ball is drawn from each bag, then find the probability that both are Red?

1. $$\frac{12}{43}$$
2. $$\frac{17}{67}$$
3. $$\frac{15}{77}$$
4. $$\frac{13}{57}$$

Answer: (c) $$\frac{15}{77}$$

Solution: Probability of getting both of Red colours, $$P = \frac{3}{7} \times \frac{5}{11}$$ $$P = \frac{15}{77}$$

1. A bag contains $$5$$ white and $$8$$ Black balls. If three balls from the bag are drawn at random, then what is the probability that they are all white?

1. $$\frac{5}{141}$$
2. $$\frac{141}{5}$$
3. $$\frac{5}{143}$$
4. $$\frac{143}{5}$$

Answer: (c) $$\frac{5}{143}$$

Solution: total number of White balls = $$5$$

number of Black balls = $$8$$, then $$P = \frac{^5C_3}{^{13}C_3}....(1)$$ where as, $$^5C_3 = \frac{5 \times 4 \times 3}{3} = 20$$ $$^{13}C_3 = \frac{13 \times 12 \times 11}{3} = 572$$ by putting these values in equation $$(1)$$, we get $$P = \frac{20}{572} = \frac{5}{143}$$